Search any question & find its solution
Question:
Answered & Verified by Expert
A body is projected vertically upwards from earth's surface of radius ' $R$ ' with velocity equal to $\frac{1}{3}$ of escape velocity. The maximum height reached by the body is
Options:
Solution:
2827 Upvotes
Verified Answer
The correct answer is:
$\frac{R}{8}$
$\Delta$ K.E. $=\Delta U$
Let mass of the particle be $\mathrm{M}$ and that of the Earth be $\mathrm{M}_{\mathrm{e}}$
$$
\therefore \quad \frac{1}{2} \mathrm{Mv}^2=\mathrm{GM}_{\mathrm{e}} \mathrm{M}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right)
$$
Also, $\mathrm{g}=\frac{\mathrm{GM}_e}{\mathrm{R}^2}$
Equation (i) can be written as,
$$
\begin{array}{ll}
& \frac{1}{2} \mathrm{v}^2=\mathrm{GM}_{\mathrm{e}}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right]=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\left[\frac{\mathrm{Rh}}{(\mathrm{R}+\mathrm{h})}\right] \\
\therefore \quad & \frac{1}{2}\left(\frac{1}{3} \mathrm{v}_{\mathrm{e}}\right)^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{1}{2}\left(\frac{1}{3} \sqrt{2 \mathrm{gR}}\right)^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{1}{2} \times \frac{1}{9}(2 \mathrm{gR})=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}=\frac{1}{9} \\
\therefore \quad & 9 \mathrm{~h}=\mathrm{R}+\mathrm{h} \\
\therefore \quad & 8 \mathrm{~h}=\mathrm{R} \\
\therefore \quad & \mathrm{h}=\frac{\mathrm{R}}{8}
\end{array}
$$
$\begin{aligned} & \text { Given } \mathrm{V}=\frac{\mathrm{Ve}}{3} \\ & \therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left[\frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{e}} / 3}\right]^2-1}=\frac{\mathrm{R}}{9-1}=\frac{\mathrm{R}}{8}\end{aligned}$
Let mass of the particle be $\mathrm{M}$ and that of the Earth be $\mathrm{M}_{\mathrm{e}}$
$$
\therefore \quad \frac{1}{2} \mathrm{Mv}^2=\mathrm{GM}_{\mathrm{e}} \mathrm{M}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}+\mathrm{h}}\right)
$$
Also, $\mathrm{g}=\frac{\mathrm{GM}_e}{\mathrm{R}^2}$
Equation (i) can be written as,
$$
\begin{array}{ll}
& \frac{1}{2} \mathrm{v}^2=\mathrm{GM}_{\mathrm{e}}\left[\frac{\mathrm{R}+\mathrm{h}-\mathrm{R}}{\mathrm{R}(\mathrm{R}+\mathrm{h})}\right]=\frac{\mathrm{GM}_{\mathrm{e}}}{\mathrm{R}^2}\left[\frac{\mathrm{Rh}}{(\mathrm{R}+\mathrm{h})}\right] \\
\therefore \quad & \frac{1}{2}\left(\frac{1}{3} \mathrm{v}_{\mathrm{e}}\right)^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{1}{2}\left(\frac{1}{3} \sqrt{2 \mathrm{gR}}\right)^2=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{1}{2} \times \frac{1}{9}(2 \mathrm{gR})=\frac{\mathrm{gRh}}{\mathrm{R}+\mathrm{h}} \\
\therefore \quad & \frac{\mathrm{h}}{\mathrm{R}+\mathrm{h}}=\frac{1}{9} \\
\therefore \quad & 9 \mathrm{~h}=\mathrm{R}+\mathrm{h} \\
\therefore \quad & 8 \mathrm{~h}=\mathrm{R} \\
\therefore \quad & \mathrm{h}=\frac{\mathrm{R}}{8}
\end{array}
$$
$\begin{aligned} & \text { Given } \mathrm{V}=\frac{\mathrm{Ve}}{3} \\ & \therefore \quad \mathrm{h}=\frac{\mathrm{R}}{\left[\frac{\mathrm{V}_{\mathrm{e}}}{\mathrm{V}_{\mathrm{e}} / 3}\right]^2-1}=\frac{\mathrm{R}}{9-1}=\frac{\mathrm{R}}{8}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.